A. Assume that the equation to be x^(½)=-4·x^(¼) ··· Eq.1
If x=0, x^(½)=0, and -4·x^(¼)=0, so that Eq.1 holds. x=0 will be one of the answers.
If x≠0, let x^(½)=t. So x^(¼)=√t, and Eq.1 can be restated as follows: t=-4√t ··· Eq.2
Divide both sides of Eq.2 with √t. We have: √t=-4 ··· Eq.3
Since √t is negative(-4), we suppose t would be the square of negative real numbers.** So that, to solve this problem, we use the rules of imaginary numbers such as i=√-1 or i²=-1.
Square both sides of Eq.3 We have: t=(-4)²=(i²·4)²=(-4)·(-4)
Therefore, x=t²=(i²·4)²·(i²·4)²=(i²·16i²)(i²·16i²)=i²·i²·(16i²)·(16i²)=(-1)·(-1)·(-16)·(-16)=(-16)·(-16)
That is: x=(-16)·(-16)
CK: From Eq.1, LHS=x^(½)=t=(-4)·(-4), RHS=-4^(¼)=-4√t=(-4)·(-4) So, LHS=RHS CKD.
Therefore, the answer is: x=0, or (-16)·(-16)
** If t is delined to be the product of 2 negative real numbers, there must be other form of combinations for x, e.g. x=(-1)(-256), (-4)·(-64), etc.
B. Assume that the equation to be x^(½)=-4·(¼)=-1 That is: √x=-1 ··· Eq.4
With the same reasons mentioned above, this question is the one on imaginary numbers as well.
Square both sides of Eq.4 to undo the root. We have: x=(-1)²=(i²)·(i²)=(-1)·(-1)
The answer is: x=(-1)·(-1)